Exercise Solutions Introduction

Slide 13

1. list can store different data types
2. data frames can store different values with different types
3. class()
4. is.character()

Slide 14

• Positive integers select elements at specific positions,
• negative integers drop elements;
• logical vectors keep elements at positions corresponding to TRUE;
• character vectors select elements with matching names.
• [] = always returns object of same class
• [[]] = can extract one element from list or data frame
• list_a <- list(a = 2, b = "hallo")
  • list_a["a"]
  • list_a[["a"]]
  • list_a$a

• [ selects sub-lists. It always returns a list; if you use it with a single positive integer, it returns a list of length one.
• [[ selects an element within a list.
• $ is a convenient shorthand: x$y is equivalent to x[["y"]].

• Use drop = FALSE if you are subsetting a matrix, array, or data frame and you want to preserve the original dimensions. You should almost always use it when subsetting inside a function.

Slide 15

1. • if() works with scalars; ifelse() works with vectors.
2. • when x is TRUE, y will be 3;
   • when FALSE, y will be NULL;
   • when NA the if statement will throw an error.

• You can rewrite any for loop to use while instead, and you can rewrite any while loop to use repeat, but the converses are not true. That means while is more flexible than for, and repeat is more flexible than while.

if("2") 3
if(2) 3

for(i in 1:3){
  print(i)
}

i <- 1
while(i < 4){
  print(i)
  i <- i + 1
}

i <- 1
repeat{
  print(i)
  i <- i + 1 
  if(i > 3) break
  
}

Slide 16

sq_root <- function(x){
  sqrt(x)
}

Exercise slides at the End

1. Combine all elements of df_list to one data frame. The result should be only a single line of code.

• do.call(rbind.data.frame, df_list)
  • do.call pass elements of df_list as arguments to rbind. It’s equivalent of rbind(df_list[[1]], df_list[[2]], df_list[[3]], ....., df_list[[length of df_list]]).

2. Fix each of the following common data frame subsetting errors:

• mtcars[mtcars$cyl == 4, ]
• mtcars[-c(1:4), ]
• mtcars[mtcars$cyl <= 5, ]
• mtcars[mtcars$cyl == 4 | mtcars$cyl == 6, ]

3. What does df[is.na(df)] <- 0 do

• replaces all NAs with 0 but keeps the dimensions.

4. How would you randomly permute the columns of a data frame?

• df[sample(1:nrow(df)), ]

5. Write a while and a repeat loop doing the same as

for(i in LETTERS[1:10]){
  print(i)
}

## [1] "A"
## [1] "B"
## [1] "C"
## [1] "D"
## [1] "E"
## [1] "F"
## [1] "G"
## [1] "H"
## [1] "I"
## [1] "J"

# while 
i <- 0
while (i < 10) {
  i = i + 1
  print(LETTERS[i])
  
}

## [1] "A"
## [1] "B"
## [1] "C"
## [1] "D"
## [1] "E"
## [1] "F"
## [1] "G"
## [1] "H"
## [1] "I"
## [1] "J"

# alternative 
i <- 1
while (i < 11) {
  print(LETTERS[i])
  i = i + 1 
}

## [1] "A"
## [1] "B"
## [1] "C"
## [1] "D"
## [1] "E"
## [1] "F"
## [1] "G"
## [1] "H"
## [1] "I"
## [1] "J"

# repeat

i <- 0
repeat{
  i = i + 1
  print(LETTERS[i])
  if (i > 9) {
    break
  }
  
}

## [1] "A"
## [1] "B"
## [1] "C"
## [1] "D"
## [1] "E"
## [1] "F"
## [1] "G"
## [1] "H"
## [1] "I"
## [1] "J"

# alternative

i <- 1
repeat{
  print(LETTERS[i])
  i = i + 1 
  if (i > 10) {
    break
  }
}

## [1] "A"
## [1] "B"
## [1] "C"
## [1] "D"
## [1] "E"
## [1] "F"
## [1] "G"
## [1] "H"
## [1] "I"
## [1] "J"

6. Write a for loop doing the same as

count <- 0
repeat{
  x     <- sample(1:6, 1)
  count <- count + 1 
  if(x == 6) break
}
print(count)

## [1] 4

for (i in 1:100) {
  x     <- sample(1:6, 1)
  count <- i
  if(x == 6) break
}
print(count)

## [1] 5

• note that \(n\) has to be high enough
• umstellen auf 95% sicherheit

7. What is the problem here?

• Curly braces are open and closed before the statement is past to the function

8. What is the problem here? Can you debug it?

• df will not be past to the final call and fast_lm is the wrong call

df <- mtcars

lin_mod <- function(df){
lm(y ~ . ,data = df)
}

simple_lin_mod <- function(data, y, x){
df <- data[ ,c(y, x)]
names(df) <- c("y", "x")
lin_mod(df)
}

simple_lin_mod(df, "mpg", "cyl")

## 
## Call:
## lm(formula = y ~ ., data = df)
## 
## Coefficients:
## (Intercept)            x  
##      37.885       -2.876

new

1. Hallo
2. World
3. Germany

Hallo world!

6. England

hist(rnorm(1000))